Aloha :)
$$a_n=\frac{2^n}{n^{\sqrt2}}\implies \frac{1}{\sqrt[n]{|a_n|}}=\sqrt[n]{\frac{n^{\sqrt2}}{2^n}}=\left(\frac{n^{\sqrt2}}{2^n}\right)^{\frac{1}{n}}=\frac{n^{\frac{\sqrt2}{n}}}{2}=\frac{(\sqrt[n]{n})^{\sqrt2}}{2}\to\frac{1}{2}$$Der Konvergenzradius ist daher \(|r|<\frac{1}{2}\).
$$b_k=k!\cdot e^{-k}\implies\left|\frac{b_k}{b_{k+1}}\right|=\frac{k!\cdot e^{-k}}{(k+1)!\cdot e^{-(k+1)}}=\frac{e}{k+1}\to 0$$Der Konvergenzradius ist daher \(r=0\).